Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <complex>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <iomanip>
#include <assert.h>
#include <array>
#include <cstdio>
#include <cstring>
#include <random>
#include <functional>
#include <numeric>
#include <bitset>
#include <fstream>

using namespace std;

struct before_main{before_main(){cin.tie(0); ios::sync_with_stdio(false);}} before_main;

#define REP(i,a,b) for(int i=a;i<(int)b;i++)
#define rep(i,n) REP(i,0,n)
#define all(c) (c).begin(), (c).end()
#define zero(a) memset(a, 0, sizeof a)
#define minus(a) memset(a, -1, sizeof a)
#define watch(a) { cout << #a << " = " << a << endl; }
template<class T1, class T2> inline bool minimize(T1 &a, T2 b) { return b < a && (a = b, 1); }
template<class T1, class T2> inline bool maximize(T1 &a, T2 b) { return a < b && (a = b, 1); }

typedef long long ll;
int const inf = 1<<29;

struct state {
  int cost, use, pos;
  state(int c, int u, int p) : cost(c), use(u), pos(p){}
};

vector<int> bfs(vector<vector<pair<int, bool>>>& G, int s, int t, int& shortestLen) {
  vector<int> dist(G.size(), inf), d2cnt(G.size()+1);
  dist[s] = 0;

  queue<state> q;
  q.emplace(0, 0, s);

  while(!q.empty()) {
    const state s = q.front(); q.pop();
    d2cnt[s.cost]++;
    for(auto& e: G[s.pos]) {
      const int next = e.first;
      if(dist[next] > s.cost + 1) {
        dist[next] = s.cost + 1;
        q.emplace(dist[next], 0, next);
      }
    }
  }

  shortestLen = dist[t];
  return d2cnt;
}

int main() {

  int N, M, s, t; cin >> N >> M >> s >> t; s--, t--;
  vector<vector<pair<int, bool>>> G(N);
  rep(i, M) {
    int x, y; cin >> x >> y;
    G[x-1].emplace_back(y-1, false);
    G[y-1].emplace_back(x-1, false);
  }

  int shortestLen;
  auto d2cnt1 = bfs(G, s, t, shortestLen);
  auto d2cnt2 = bfs(G, t, s, shortestLen);

  ll ans = 0;

  rep(i, N) {
    if(shortestLen-i-2 >= 0)
      ans += (ll)d2cnt1[i] * d2cnt2[shortestLen-i-2];
  }

  cout << ans << endl;

  return 0;
}